An elevator with a man inside, of gross mass 500KG, starts moving upwards with a costant acceleration and acquires a velocity os 2m/s after travelling a distance os 3m.Find the pull in the cables during the accelerated motion.

Question

An elevator with a man inside, of gross mass 500KG, starts moving upwards with a costant acceleration and acquires a velocity os 2m/s after travelling a distance os 3m.Find the pull in the cables during the accelerated motion.

If the elevator, before stopping, moves with a constant deceleration from a constant velocity of 2m/s and comes to rest in 2 second, calculate the force exerted by the man of 75kg mass on the floor on the elevator.

Answer 1

To answer the first question about the pull in the cables during the accelerated motion, we can start by applying Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = m * a).

In this case, the elevator has a mass of 500 kg, and it acquires a velocity of 2 m/s after traveling a distance of 3 m. We need to determine the acceleration first.

Using the kinematic equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can rearrange the equation to solve for the acceleration:

a = (v^2 - u^2) / (2s)
= (2^2 - 0^2) / (2 * 3)
= 4 / 6
= 2/3 m/s^2

Now that we have the acceleration, we can calculate the net force acting on the elevator during the accelerated motion:

F = m * a
= 500 kg * (2/3 m/s^2)
= 1000/3 N

Therefore, the pull in the cables during the accelerated motion is approximately 333.33 N.

Moving on to the second question about the force exerted by the man on the elevator floor during the deceleration phase, we can use the same approach.

Given that the elevator is moving with a constant velocity of 2 m/s and comes to rest in 2 seconds, we need to find the acceleration during this deceleration phase. The formula for acceleration in this case is:

a = (v - u) / t
= (0 - 2) / 2
= -2/2
= -1 m/s^2

Since the acceleration is negative (indicating deceleration), the force exerted by the man on the elevator floor can be calculated using the formula:

F = m * |a|
= 75 kg * |-1 m/s^2|
= 75 N

Therefore, the force exerted by the man on the elevator floor during the deceleration phase is 75 N.

An elevator with a man inside, of gross mass 500KG, starts moving upwards with a costant acceleration and acquires a velocity os 2m/s after travelling a distance os 3m.Find the pull in the cables during the accelerated motion.

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To answer the first question about the pull in the cables during the accelerated motion, we can start by applying Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = m * a).